Question

Find the sum of the following series to n terms : (1-7) $1^3+3^3+5^3+7^3+....$

Answer 1

$1^3+3^3+5^3+7^3+....Let{T}_{n}bethenthtermofthisseriesthen,T_n=[1+(n-1)2{]}^3=(2n-1{)}^3=(2n{)}^3-3(2n{)}^2.1+{3.1}^2.2n-1^3=8n^3-12n^2+6n-1[\because (a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3]={\sum }_{k-1}^n{T}_k={\sum }_{k-1}^n(8k^3-12k^2+6k-1)=8{\sum }_{k-1}^n{k}^3-12{\sum }_{k-1}^n{k}^2+6{\sum }_{k-1}^{n}k-{\sum }_{k-1}^{n}1=8\left[\frac{n(n+1{)}^2}{2}\right]-12\left[\frac{n(n+1)(2n+1)}{6}\right]+6\left[\frac{n(n+1)}{2}\right]-n=8\frac{n^2(n+1{)}^2}{4}-\frac{12\left[n\right(n+1\left)\right(2n+1\left)\right]}{6}+3\left[n\right(n+1\left)\right]-n=2n^2(n+1{)}^2-2n(n+1\left)\right(2n+1)+3(n+1)n-n=n(n+1\left)\right[2n^2(n+1)-2n(2n+1)+3n]-n=(n+1\left)\right[2n^3+2n^2-4n^2-2n+3n]-n=(n+1\left)\right[2n^3-2n^2+n]-n=2n^4-2n^3+n^2+2n^3-2n^2+n-n=2n^4-n^2=n^2(2n^2-1)\therefore 1^3+3^3+5^3+7^3+....tonterms=n^2(2n^2-1)$